I have been asked to find:
$\lim_{x\to 0} \, \cos \left(\frac{\pi -\pi \cos ^2(x)}{x^2}\right)$$=-1$
Without using l'Hôpital's rule. But I have no idea how to due it, could someone show a step by step process all the way to the answer?
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Hint: We have $$\frac{\pi -\pi\cos^2 x}{x^2}=\frac{\pi\sin^2 x}{x^2}=\pi \left(\frac{\sin x}{x}\right)^2.$$
André Nicolas
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I do not understand why (sinx)/x has limit of one as x approaches zero because sin[0]=1 and when the denominator approaches zero this number should be infinite large what am i missing? – ALEXANDER Jul 30 '14 at 02:32
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The fact that $\lim_{x\to 0}\frac{\sin x}{x}=1$ is proved in calculus courses, usually geometrically, when one needs the derivative of $\sin x$, for this limit is a critical part of the proof. Note that $\sin 0=0$. If you look back to the differentiation of trigonometric functions part of your book, you will see this limit prominently discussed. – André Nicolas Jul 30 '14 at 02:50
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Unfortunately my book only refers to a proof by the squeeze theorem, but It does not show the proof, would you be able to show me or should I pose a new question? – ALEXANDER Jul 30 '14 at 02:59
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It has been done several times on MSE. Yes, it is a squeezing argument. The proof is in a great many places. If you ask the question on MSE, it will probably be quickly closed as a duplicate, but you will have at least one reference. I cannot show you in comments, it really requires a picture. – André Nicolas Jul 30 '14 at 03:19
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You are welcome. I would find it on MSE, but I am not good at searching. I am surprised it is not in your calculus book, it has been in every calculus book I have ever used as a text (many!). – André Nicolas Jul 30 '14 at 05:18
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Notice that:
$$\frac{\pi -\pi\cos^2 x}{x^2}=\frac{\pi\sin^2 x}{x^2}=\pi \left(\frac{\sin x}{x}\right)^2.$$
As you know, the limit:
$$\lim_{x\to 0} \, \left(\frac{\sin x}{x}\right)=1$$
Therefore, our answer is:
$$\pi*1 = \pi$$
And
$$\cos(\pi) = -1$$
Varun Iyer
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